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If the roots of the equation $\left(p^2+q^2\right) x^2-2 q(p+r) x+\left(q^2+r^2\right)=0$ be real and equal, then $p, q, r$ will be in
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$G.P$
Given equation is
$\left(p^2+q^2\right) x^2-2 q(p+r) x+\left(q^2+r^2\right)=0$
Roots are real and equal, then
$4 q^2(p+r)^2-4\left(p^2+q^2\right)\left(q^2+r^2\right)=0$
$\Rightarrow q^2\left(p^2+r^2+2 p r\right)-\left(p^2 q^2+p^2 r^2+q^4+q^2 r^2\right)=0$
$\Rightarrow q^2 p^2+q^2 r^2+2 p q^2 r-p^2 q^2-p^2 r^2-q^4-q^2 r^2=0$
$\Rightarrow 2 p q^2 r-p^2 r^2-q^4=0 \Rightarrow\left(q^2-p r\right)^2=0$
Hence $q^2=p r$. Thus $p, q . r$ in G.P.
$\left(p^2+q^2\right) x^2-2 q(p+r) x+\left(q^2+r^2\right)=0$
Roots are real and equal, then
$4 q^2(p+r)^2-4\left(p^2+q^2\right)\left(q^2+r^2\right)=0$
$\Rightarrow q^2\left(p^2+r^2+2 p r\right)-\left(p^2 q^2+p^2 r^2+q^4+q^2 r^2\right)=0$
$\Rightarrow q^2 p^2+q^2 r^2+2 p q^2 r-p^2 q^2-p^2 r^2-q^4-q^2 r^2=0$
$\Rightarrow 2 p q^2 r-p^2 r^2-q^4=0 \Rightarrow\left(q^2-p r\right)^2=0$
Hence $q^2=p r$. Thus $p, q . r$ in G.P.
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